(a) If [tex]y(t)[/tex] is the mass after [tex]t[/tex] days and [tex]y(0) = A[/tex] then [tex]y(t) = Ae^{kt}.[/tex]
[tex]y(1) = Ae^k = 0.945A \implies e^k = 0.945 \implies k = \ln 0.945[/tex]
[tex]\text{Then } Ae^{(\ln 0.945)t} = \frac{1}{2}A\ \Leftrightarrow\ \ln e^{(\ln 0.945)t} = \ln \frac{1}{2}\ \Leftrightarrow\ ( \ln 0.945) t = \ln \frac{1}{2} \Leftrightarrow \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 12.25 \text{ years}[/tex]
(b)
[tex]Ae^{(\ln 0.945) t} = 0.20 A \ \Leftrightarrow\ (\ln 0.945) t \ln \frac{1}{5}\ \implies \\ \\
t = - \frac{\ln 5}{\ln 0.945} \approx 28.45\text{ years}[/tex]